Jacobisymbols modulo $pq$, both prime I am trying to find the number of $a$'s for which $$\left( \frac{a}{pq} \right) = 1$$ with $p$ and $q$ both prime, but I am not sure how to approach this problem. I was using the identity with the Legendre symbols $$\left( \frac{a}{pq} \right) = \left( \frac{a}{p} \right) \left( \frac{a}{q} \right)$$ because $p$ and $q$ are both prime. I would like to check whether $\left( \frac{a}{p} \right) = 1$ and $\left( \frac{a}{q} \right) = 1$ and whether $\left( \frac{a}{p} \right) = -1$ and $\left( \frac{a}{q} \right) = -1$, but then I am stuck here, since I don't know anything about the $a$. I could distinguish in $a$ odd or $a$ even, but it won't be much easier. Can somebody please help me understanding this question?
I think the answer has to be $$\frac{(p-1)(q-1)}{2}$$ since we delete the $0$ and then there will be intuitively half of all possibilities having the same sign.
 A: I assume that $p,q$ are odd distinct primes. We want to determine the number of $a$ (in $\{0,\dots,pq-1\}$ I assume) with $\left(\frac{a}{pq}\right)=1$. Note that this is equivalent to $\left(\frac{a}{p}\right)=\left(\frac{a}{q}\right)$ and $a\nmid p$.  For every $a_1\in \{1,\dots,p-1\}$ there are exactly $\frac{q-1}{2}$ numbers $a_2\in\{0,\dots,q-1\}$ such that $\left(\frac{a_1}{p}\right)=\left(\frac{a_2}{q}\right)$ (because exactly have of the residue classes are quadratic (non)-residues modulo a prime), so exactly $(p-1)\frac{q-1}{2}$ tuples $(a_1,a_2)$ from $\{0,\dots,p-1\}\times\{0,\dots,q-1\}$ satisfy $p\nmid a_1$ and $\left(\frac{a_1}{p}\right)=\left(\frac{a_2}{q}\right)$.
Using the isomorphism $\Bbb Z/pq\cong\Bbb Z/p\times \Bbb Z/q$ (which preserves the Legendre/Jacobi symbol as they only depend on the residue class) from the Chinese remainder theorem we get that the desired number is $\frac{(p-1)(q-1)}{2}$ as you predicted.
Note that if $p,q$ are not distinct, then the number is $p^2-p$ as $\left(\frac{a}{pq}\right)=\left(\frac{a}{p^2}\right)=\left(\frac{a}{p}\right)^2=1$ iff $p\nmid a$.
